A Note on Nested Words Andreas Blass¤ Yuri Gurevichy Abstract For every regular language of nested words, the underlying strings form a context-free language, and every context-free language can be obtained in this way. Nested words and nested-word automata are gen- eralized to motley words and motley-word automata. Every motley- word automation is equivalent to a deterministic one. For every reg- ular language of motley words, the underlying strings form a ¯nite intersection of context-free languages, and every ¯nite intersection of context-free languages can be obtained in this way. 1 Introduction In [1], Rajeev Alur and P. Madhusudan introduced and studied nested words, nested-word automata and regular languages of nested words. A nested word is a string of letters endowed with a so-called nested relation. Alur and Madhusudan prove that every nested-word automaton is equivalent to a deterministic one. In Section 2, we recall some of their de¯nitions. We quote from the introduction to [1]. \The motivating application area for our results has been soft- ware veri¯cation. Given a sequential program P with stack-based control flow, the execution of P is modeled as a nested word with nesting edges from calls to returns. Speci¯cation of the program is given as a nested word automaton A, and veri¯cation corre- sponds to checking whether every nested word generated by P is ¤Math Dept, University of Michigan, Ann Arbor, MI 48109; [email protected] yMicrosoft Research, Redmond, WA 98052; [email protected] 1 accepted by A. Nested-word automata can express a variety of requirements such as stack-inspection properties, pre-post condi- tions, and interprocedural data-flow properties." In Section 3, we show that all context-free properties, and only context- free properties, can be captured by a nested-word automaton in the following precise sense. For every regular language of nested words, the underlying strings form a context-free language, and every context-free language can be obtained in this way. We continue the quotation from [1]. \If we were to model program executions as words, all of these properties are non-regular, and hence inexpressible in classical speci¯cation languages based on temporal logics, automata, and ¯xpoint calculi (recall that context-free languages cannot be used as speci¯cation languages due to nonclosure under intersection and undecidability of key decision problems such as language in- clusion)." Intersections of context-free languages naturally arise in applications. Think of multi-threaded programs for example. In Section 4, we general- ize nested words to motley words: strings with several nested relations. We introduce motley-word automata and use the nested-word-automaton deter- minization procedure to show that every motley-word automaton is equiva- lent to a deterministic one. We show that every ¯nite intersection (that is the intersections of ¯nitely many) of context-free languages, and only such intersections, can be captured by a motley-word automaton in the following precise sense. For every regular (that is accepted by a motley-word automa- ton) language of motley words, the language of underlying strings is a ¯nite intersection of context-free languages, and every intersection of context-free languages can be obtained in this way. 2 2 Preliminaries To make this note self-contained, we recapitulate here some de¯nitions from [1]. 2.1 Nested words A nested relation º over f1; 2; : : : ; kg is a binary relation satisfying the fol- lowing condition: if º(i; i0) and º(j; j0) and i · j then either i < i0 < j < j0 or else i < j < j0 < i0. If º(i; j) then i is a call position and the call-predecessor of j, and j is a return position and the return-successor of i. A nested word over an alphabet § is a pair (a1 : : : ak; º) where each ai 2 § and º is a nested relation over f1; 2; : : : ; kg. A position i that is neither a call position nor a return position is an internal position of w. NW(§) is the set of nested §-words. A language of nested §-words is a subset of NW(§). 2.2 NW automata An nw automaton A over an alphabet § is a quadruple (Q; Qin; ±; Qf ) similar to a usual nondeterministic automaton with set Q of states, set Qin ⊆ Q of initial states and set Qf ⊆ Q of ¯nal states, except that ± is not a single transition relation but a triple h±c; ±i; ±ri where ² ±c ⊆ Q £ § £ Q is the transition relation for call positions, ² ±i ⊆ Q £ § £ Q is the transition relation for internal positions, and ² ±r ⊆ Q £ Q £ § £ Q is the transition relation for return positions. The automaton A is deterministic if it satis¯es the following conditions: - There is exactly one initial state. - ±c is the relational form of a function from Q £ § to Q. In other words, 0 0 if (p; a; q) and (p; a; q ) belong to ±c then q = q . - ±i is the relational form of a function from Q £ § to Q. - ±c is the relational form of a function from Q £ Q £ § to Q. 3 A run of A over an nested word (a1 : : : ak; º) is a sequence q0; q1; : : : ; qk of states such that q0 is an initial state, if i is a call position then (qi¡1; ai; qi) 2 ±c, if i is an internal position then (qi¡1; ai; qi) 2 ±i, and if i is a return position with call-predecessor j then (qi¡1; qj¡1; ai; qi) 2 ±r. The run q0; q1; : : : ; qk is accepting if qk is a ¯nal state. A accepts a word w if it has an accepting run on w. The language L(A) of A is the set of nested words accepted by A. A language (that is a set) of nested words is regular if there is an nw automaton A such that L = L(A). Two nw automata A; B are equivalent if L(A) = L(B). Alur and Madhusudan prove that every nw automaton A is equivalent to a deterministic nw automaton [1, Theorem 1]. 3 Nested-Word Automata and Context-Free Languages In this section, we show that, ² for every regular language of nested words, the underlying strings form a context-free language, and ² every context-free language can be obtained in this way. De¯nition 1. A is call explicit if ±c and ±i are disjoint. Lemma 2. For every nested-word automaton A = (Q; Qin; ±; Qf ), there is a call-explicit nw automaton A0 such that L(A0) = L(A). Furthermore, if A is deterministic then so is A0. 0 0 0 0 0 0 0 0 Proof. The desired A = (Q ;Qin; ± ;Qf ) where Q , Qin and Qf are Q£f0; 1g, Qin £ f0g and Qf £ f0g respectively. Further: 0 ±c = f((p; d); a; (q; 1)) : (p; a; q) 2 ±c ^ d 2 f0; 1gg 0 ±i = f((p; d); a; (q; 0)) : (p; a; q) 2 ±i ^ d 2 f0; 1gg 0 ±r = f((p1; d1); (p2; d2); a; (q; 0)) : (p1; p2; a; q) 2 ±r ^ d1; d2 2 f0; 1gg 0 0 0 Clearly ±c and ±i are disjoint, and A is deterministic if A is so. Every accepting run q0; : : : ; qk of A on (a1 : : : ak; º) gives rise to an accepting run (q0; d0);:::; (qk; dk) where di = 1 if and only if i is a call position. And 0 every accepting run (q0; d0);:::; (qk; dk) of A on (a1 : : : ak; º) gives rise to an accepting run q0; : : : ; qk of A. 4 De¯nition 3. The projection P (w) of a nested word w = (x; º) is the string x. The projection P (L) of a language L of nested words is the language fP (w): w 2 Lg. Proposition 4. The projection of a regular nw language is context-free. Proof. Let L = L(A) where A is an nw automaton (Q; Qin; ±; Qf ). We con- struct a (nondeterministic) pushdown automaton B with L(B) = P (L) that accepts on empty stack and ¯nal state. The sets of states, initial states and ¯nal states of B are Q, Qin and Qf respectively. The stack alphabet of B is Q[f?g where ? is the bottom-of-the-stack symbol. The transition function ¢ = ¢c [ ¢i [ ¢r where the components are as follows. 0 0 ¢i = f(p; a; p ; q):(p; a; q) 2 ±i ^ p 2 Qg 0 0 ¢c = f(p; a; p ; q; +p):(p; a; q) 2 ±c ^ p 2 Qg 0 0 0 ¢r = f(p; a; p ; q; ¡):(p; p ; a; q) 2 ±r ^ p 2 Qg The intended meaning of an instruction (p; a; p0; q) is this: if the current state is p, the input symbol is a and the top stack symbol is p0 then to go state q (and move one-letter to the right on the input word). The additional \+p" means: push p onto the stack. The additional \¡" means: pop the stack. An accepting run q0; : : : ; qk of A on (a1 : : : a`; º) gives rise to an accepting run (q0;U0) ::: (qk;Uk) of B on a1 : : : ak where U0;:::;Uk are stack contents de¯ned inductively. U0 = ?. Let i > 0. If i is an internal position then Ui = Ui¡1. If i is a call position then Ui = qi¡1Ui¡1. Suppose that u is a return position with call-predecessor j. The number of call positions < i exceeds the number of return positions < i, and so Ui¡1 6= ?. Ui is obtained from Ui¡1 by popping the top symbol.